Advertisements
Advertisements
Question
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
Advertisements
Solution
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(bb(cos^2θ) + bb(sin^2θ))/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ...............[sin2θ + cos2θ = 1]
= `1/sinθ xx 1/bbcosθ`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`(1 + cos theta + sin theta)/(1 + cos theta - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - x^2/c^2 = 1`
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
(1 + sin A)(1 – sin A) is equal to ______.
